Problem: Multiply the following complex numbers: $({-4+2i}) \cdot ({4-4i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4+2i}) \cdot ({4-4i}) = $ $ ({-4} \cdot {4}) + ({-4} \cdot {-4}i) + ({2}i \cdot {4}) + ({2}i \cdot {-4}i) $ Then simplify the terms: $ (-16) + (16i) + (8i) + (-8 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -16 + (16 + 8)i - 8i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -16 + (16 + 8)i - (-8) $ The result is simplified: $ (-16 + 8) + (24i) = -8+24i $